Answer:
3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate.
Explanation:
The balance chemical equation is :
Mass of barium hydroxide octahydrate = 6.5 g
Moles of barium hydroxide octahydrate = 
According to reaction, 2 moles of ammonium thiocyanate reacts with1 mole of barium hydroxide octahydrate. The 0.020635 moles of barium hydroxide octahydrate will react with:
Mass of 0.04127 moles of ammonium thiocyanate;
3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate
The net force is 11.1 because 23.5-12.4 is your net force
hope this helps:)
Answer:
The answer is
<h2>0.89 atm </h2>
Explanation:
To convert from kPa to atm we use the conversion
101.325 kPa = 1 atm
If
101.325 kPa = 1 atm
Then
90.23 kPa will be
We have the final answer as
<h3>0.89 atm</h3>
Hope this helps you
The relative molecular mass of acid A : 50 g/mol
<h3>Further explanation</h3>
Given
40.0 cm³(40 ml) of 0.2M sodium hydroxide
0.2g of a dibasic acid
Required
the relative molecular mass of acid A
Solution
Titration formula
M₁V₁n₁=M₂V₂n₂
n=acid/base valence(number of H⁺/OH⁻)
NaOH ⇒ n = 1
Dibasic acid = diprotic acid (such as H₂SO₄)⇒ n = 2
mol = M x V
Input the value in the formula :(1 = NaOH, 2=dibasic acid)
0.2 x 40 x 1 = M₂ x V₂ x 2
M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A
The relative molecular mass of acid A (M) :
Answer:
0.287 mole of PCl5.
Explanation:
We'll begin by calculating the number of mole in 51g of Cl2. This is illustrated below:
Molar mass of Cl2 = 2 x 35.5 = 71g/mol
Mass of Cl2 = 51g
Number of mole of Cl2 =..?
Mole = Mass /Molar Mass
Number of mole of Cl2 = 51/71 = 0.718 mole
Next, we shall write the balanced equation for the reaction. This is given below:
P4 + 10Cl2 → 4PCl5
Finally, we determine the number of mole of PCl5 produced from the reaction as follow:
From the balanced equation above,
10 moles of Cl2 reacted to produce 4 moles of PCl5.
Therefore, 0.718 mole of Cl2 will react to produce = (0.718 x 4)/10 = 0.287 mole of PCl5.
Therefore, 0.287 mole of PCl5 is produced from the reaction.
